CBSE 2025, Set 2

Question

(a) Define the term 'drift velocity' of conduction electrons in a conductor.
(b) A conductor of length $l$ and area of cross-section $A$ is connected across an ideal battery of emf $V$. Derive the formula for the current density in terms of relaxation time $	au$.

Edvaya · Accepted Answer

The correct answer is Subjective. **(a) Drift Velocity:**
Drift velocity is the average velocity with which the free electrons move (drift) towards the positive end of a conductor under the influence of an external electric field.

**(b) Current Density Derivation:**
By definition, current density $j$ is the current per unit area:
$$j = \frac{I}{A}$$
We know the relation between current and drift velocity is:
$$I = neAv_d$$
Therefore, substituting $I$:
$$j = \frac{neAv_d}{A} = nev_d$$
We also know the expression for drift velocity in terms of relaxation time $	au$:
$$v_d = \frac{eE	au}{m}$$
Substituting $v_d$ into the current density equation:
$$j = n e \left( \frac{eE	au}{m} ight)$$
$$j = \frac{ne^2	au E}{m}$$
Since the electric field $E$ in a conductor of length $l$ connected to a voltage $V$ is $E = \frac{V}{l}$, we can also write:
$$\boxed{j = \frac{ne^2	au}{m} \left( \frac{V}{l} ight)}$$

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